The entire problem is as follows:
Consider $\dot{x} = y+x^2-y^2$ and $\dot{y} = -x-2xy$. (a) Find all equilibria for this system and determine their stability. (b) Find the Hamiltonian of the system. (c) Find a solution curve that connects a saddle point. (d) Does the curve obtained in (c) connect any other saddle points?
Below is my answer to parts $(a)$ and $(b)$:
a)
There are four equilibrium points here, all of which can be found using relatively simple algebra. They are $E_1 = (0,0), E_2 = (0,1), E_{3,4} = \left( \pm \frac{\sqrt{3}}{2}, - \frac{1}{2} \right)$. The stability of these is found below:
$$Df\left( x, y \right) = \left( \begin{matrix} 2x & 1-2y \\ -1-2y & -2x\end{matrix}\right)$$
$$Df\left( E_1 \right) = \left( \begin{matrix} 0 & 1 \\ -1 & 0\end{matrix}\right) \implies \lambda_{1,2} = \pm i$$
Thus $E_1$ is a center.
$$Df\left( E_2 \right) = \left( \begin{matrix} 0 & -1 \\ -3 & 0\end{matrix}\right) \implies \lambda_{1,2} = \pm \sqrt 3$$ Thus $E_2$ is a saddle.
$$Df\left( E_3 \right) = \left( \begin{matrix} \sqrt 3 & 2 \\ 0 & - \sqrt 3\end{matrix}\right) \implies \lambda_{1,2} = \pm \sqrt 3$$ Thus $E_3$ is a saddle.
$$Df\left( E_4 \right) = \left( \begin{matrix} -\sqrt{3} & 2 \\ 0 & \sqrt 3\end{matrix}\right) \implies \lambda_{1,2} = \pm \sqrt 3$$ Thus $E_4$ is a saddle.
b)
$$H(x,y) = \int (y + x^2 - y^2)dy = \frac{1}{2}y^2 + x^2 y - \frac{1}{3}y^3 + f(x)$$ Thus $$-\frac{d}{dx} H(x,y) = -\frac{d}{dx} [x^2 y + f(x)] = \dot y = -x-2xy$$ which implies $$-2xy -f'(x) = -x-2xy \implies f'(x) = x$$ thus $f(x) = \frac{1}{2}x^2$. So the Hamiltonian is $$H(x,y) = \frac{1}{2}y^2 + x^2 y - \frac{1}{3}y^3 + \frac{1}{2}x^2$$.
Now that I have all of this information however, (c) seems like a strangely vague question. Is all I have to do here plug in a saddle point and see what solution it gives me? A walkthrough of this would be helpful, and then would in turn help me with part (d). Thanks in advance for any and all advice!